Chapter 6.7, Problem 33WE

To determine

To factor : the expression $2n^3-14n+12$.

Answer to Problem 33WE

The complete factor t is $2n^3-14n+12=2\left(n+3\right){\left(n-2\right)}^2$.

Explanation of Solution

Given information:

  $(n+3)\text{is a factor of }2n^3-14n+12$.

Formula used:

  $\text{Dividend=Quotient×Divisor+Remainder}$

Calculation:

We have ,

  $(n+3)\text{ factor of (}2n^3-14n+12)$

This implies that $\text{(}2n^3-14n+12)$ can be divided by $\text{(}n+3)$.

Now, using polynomial division

Step 1.

  $n+3\begin{array}{l}\begin{array}{c}\hfill 2n^2\\ \hfill \overline{)\begin{array}{l}2n^3+0n^2-14n+12\\ 2n^3+6n^2\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $-6n^2-14n+12$

Step 2.

  $n+3\begin{array}{l}\begin{array}{c}\hfill 2n^2-6n\\ \hfill \overline{)\begin{array}{l}-6n^2-14n+12\\ -6n^2-18n\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $4n+12$

Step 3.

  $n+3\begin{array}{l}\begin{array}{c}\hfill 2n^2-6n+4\\ \hfill \overline{)\begin{array}{l}4n+12\\ 4n+12\end{array}}\end{array}\hfill \end{array}$  $\leftarrow \text{Subtract}$

  $0$

  $\begin{array}{l}=2n^3-14n+12=(n+3)(2n^2-6n+4)+0\\ =2n^3-14n+12=\left(n+3\right)\left(2n^2-6n+4\right)\end{array}$

Further,

  $=2n^2-6n+4=2n^2-4n-2n+4$

  $\begin{array}{l}=2n(n-2)-2(n-2)\\ =(2n-2)(n-2)\\ =2(n-2)(n-2)\\ $ =2{(n-2)}^2$\end{array}$

Hence, the complete factor of $\text{(}2n^3-14n+12)$ :-

  $2n^3-14n+12=2\left(n+3\right){\left(n-2\right)}^2$